DISCUSSION #5
FRIDAY MAY 4TH 2007
Sophie Engle (Teacher Assistant)
ECS20: Discrete Mathematics
Discussion Outline
2


Quick Announcements
Induction Examples
 Example 2
(page 268)
 Example 7 (page 272)
 Example 12 (page 276)
 Note
that I will be working these out on the board.
If you want to see how they are done please see the
book.
Announcements
3

Homework 3:
 Grades posted on

my.ucdavis.edu.
Homework 5:
 Due
Wednesday May 9th.
 Total of 7 problems from section 4.1.

Midterm:
 Was
one problem that several students had
problems with.
 May reappear on final exam!
Midterm Notes
4

Prove that 𝑎 ≡ 𝑏 mod 𝑚 implies that
𝑎𝑐 ≡ 𝑏𝑐 (mod 𝑚𝑐) for any integer 𝑐.



We know that 𝑎 ≡ 𝑏 mod 𝑚 means that (𝑎 – 𝑏) is a
multiple of 𝑚 so (𝑎 – 𝑏) = 𝑘𝑚 for some 𝑘.
Therefore 𝑎𝑐 – 𝑏𝑐 = (𝑎 – 𝑏)𝑐 = 𝑘𝑚𝑐 is a
multiple of 𝑚𝑐.
The fact that 𝑎𝑐 – 𝑏𝑐 is a multiple of 𝑚𝑐 is the definition
what it means for 𝑎𝑐 to be congruent to 𝑏𝑐 (mod 𝑚𝑐).
5
Mathematical Induction
Chapter 4, Section 1
Mathematical Induction
6

Step 0: Create a Conjecture
 Figure
out each step looks like based on 𝑛.
 Figure out a pattern for the results based on 𝑛.

Step 1: Basis Step
 Show that

𝑃(1) is true.
Step 2: Inductive Step
 Assume 𝑃(𝑘)
is true.
 Show that if 𝑃(𝑘) is true then 𝑃(𝑘 + 1) is true.
Example 2
7
Conjecture a formula for the sum of the first
𝑛 positive odd integers and prove your
conjecture using mathematical induction.
Example 2
8

Step 0: Create a Conjecture
 Figure
out each step looks like based on 𝑛.
 Figure out a pattern for the results based on 𝑛.
Example 2
9

Step 1: Basis Step
 Show that
𝑃(1) is true.
𝑃(1) = 12 = 1 is true since the sum of the
first one positive odd integer is 1.
(Notice the last term is 2𝑛 – 1 = 2 – 1 = 1.
Therefore we start and end with 1.)
Example 2
10

Step 2: Inductive Step
 Assume 𝑃(𝑘)
is true.
 Show that if 𝑃(𝑘) is true then 𝑃(𝑘 + 1) is true.
Homework 5
11

For your homework, when proving by induction:
 Always
include the basis step.
 Always state your assumption that 𝑃(𝑘) is true.

These steps may seem unnecessary, but you do
not have a proof by induction without them.
Example 7
12

The harmonic numbers 𝐻𝑗 for 𝑗 = 1, 2, 3, …,
are defined by:
j
1 1
1
1
H j  1    
2 3
j k 1 k
Use mathematical induction to show that:
H 2n  1 
n
2
whenever 𝑛 is a nonnegative integer.
Example 7
13

Step 1: Basis Step
 Show that
𝑃(1) is true.
Example 7
14

Step 2: Inductive Step
 Assume 𝑃(𝑘)
is true.
 Show that if 𝑃(𝑘) is true then 𝑃(𝑘 + 1) is true.
Example 7
15

What does this mean?
 The harmonic series is
divergent!
 Means that limit approaches infinity.

1


n 1 n
Example 12
16

An odd number of people stand in a yard at
mutually distinct distances. At the same time
each person throws a pie at their nearest
neighbor, hitting this person.
Use mathematical induction to show that there
is at least one survivor, that is, at least one
person who is not hit by a pie.
Example 12
17

Step 0: Create a Conjecture
 We
want to show that some conjecture 𝑃(𝑛) is true
for all positive integers 𝑛.
 We can’t use 𝑛 alone to represent the number of
people in the fight, since 𝑛 may be even.
 Therefore we need to represent the number of
people based on 𝑛 such that:
 The number of people is always odd.
 There are
at least 3 people.
(a one person pie fight isn’t very interesting)
Example 12
18

Step 0: Create a Conjecture
 Let
𝑃(𝑛) be the statement that for 2𝑛 + 1 people in
the pie fight, at least one person is not hit by a pie.

Step 1: Basis Step
 For
𝑃(1) there are 2 ∗ 1 + 1 = 3 people.
 Call these three
 Since each
Xander, Yasmin, and Zane.
person is at mutually distinct distances,
for each person there is only one closest neighbor.
Example 12
19

Step 1: Basis Step
 Suppose the
shortest distance is between Xander
and Yasmin.
 Then Xander will throw a pie at Yasmin.
 Then Yasmin will throw a
 Therefore, there is
pie at Xander.
nobody to throw a pie at Zane.
 Zane will throw a pie at the closest person.
 Hence, for
𝑃(1), one person will not be hit by a pie.
Example 2
20

Step 2: Inductive Step
 Assume 𝑃(𝑘)
is true.
 Assume for 2𝑘
 Show that
+ 1 people, one person is not hit by a pie.
if 𝑃(𝑘) is true then 𝑃(𝑘 + 1) is true.
 For 𝑃(𝑘 + 1), there are
2(𝑘+1)+1 = 2𝑘 + 3 people.
 Therefore there are also 2𝑘 + 3 pies to throw.
 Let the shortest distance be between Xander & Yasmin.


Therefore, Xander throws a pie at Yasmin.
Therefore, Yasmin throws a pie at Xander.
 There are
now 2𝑘 + 1 pies to throw by 2𝑘 + 1 people.
Example 2
21
 Show that
if 𝑃(𝑘) is true then 𝑃(𝑘 + 1) is true.
 There are
now 2𝑘 + 1 pies to throw by 2𝑘 + 1 people.
 If Zane throws his pie at Xander or Yasmin then:



There are 2𝑘 pies left to throw at 2𝑘 + 1 people.
If there are less pies than people, someone must survive!
This is known as the pigeon hole principle.
 If no one throws a


pie at Xander or Yasmin then:
We now have 2𝑘 + 1 people.
By our assumption 𝑃(𝑘), for 2𝑘 + 1 people, at least one
person is not hit by a pie.